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Vikentia [17]
2 years ago
11

What is the perimeter of this quadrilateral? (5,5) (2, 4) (4, 1) (6, 1)

Mathematics
1 answer:
lbvjy [14]2 years ago
8 0

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{4}) ~\hfill AB=\sqrt{[ 2- 5]^2 + [ 4- 5]^2} \\\\\\ AB=\sqrt{(-3)^2+(-1)^2}\implies \boxed{AB=\sqrt{10}} \\\\[-0.35em] ~\dotfill\\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) ~\hfill BC=\sqrt{[ 4- 2]^2 + [ 1- 4]^2}

BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}

DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89

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Step-by-step explanation:

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4 0
2 years ago
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earnstyle [38]

Answer:

B. \frac{4x^{2}+24x+31}{3x^{2}+13x+12}

Step-by-step explanation:

We have been given a rational expression \frac{(x+5)}{(3x+4)}+\frac{(x+4)}{(x+3)} and we are asked to simplify our rational expression.

We can see that our denominators are not equal. Since the two denominators do not share any common factors, the common denominator is simply the product of these two denominators.      

To keep the value of expression same we will multiply the same quantity with numerators and denominators.

\frac{(x+5)(x+3)}{(3x+4)(x+3)}+\frac{(x+4)(3x+4)}{(x+3)(3x+4)}  

Now let us simplify our expression using FOIL.  

\frac{x^{2}+3x+5x+15}{3x^{2}+9x+4x+12}+\frac{3x^2+4x+12x+16}{3x^{2}+4x+9x+12}

\frac{x^{2}+8x+15}{3x^{2}+13x+12}+\frac{3x^2+16x+16}{3x^{2}+13x+12}    

Now we have same denominators, so we can add our numerators.

\frac{x^{2}+8x+15+3x^2+16x+16}{3x^{2}+13x+12}    

Now let us combine like terms.

\frac{(1+3)x^{2}+(8+16)x+15+16}{3x^{2}+13x+12}  

\frac{4x^{2}+24x+31}{3x^{2}+13x+12}  

Therefore, the simplest form of our given rational expression will be \frac{4x^{2}+24x+31}{3x^{2}+13x+12}  and option B is the correct choice.

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A

Step-by-step explanation:

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3 years ago
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Multiply the cost by the area
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6 0
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28 + 6 = 34 and 34 - 3 = 31!!!!!!!

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